Question

A galvanic cell is constructed that carries out the reaction

${\mathbf{\text{Pb}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{Cr}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathbf{\text{Pb}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{Cr}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

If the initial concentration of ${\mathbf{\text{Pb}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$is$\mathbf{0}\mathbf{.15}\mathbf{\text{M}}$,that of${\mathbf{\text{Cr}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$ is $\mathbf{0}\mathbf{.20}\mathbf{\text{M}}$, and that of ${\mathbf{\text{Cr}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$is ,$\mathbf{0}\mathbf{.0030}\mathbf{\text{M}}$ calculate the initial voltage generated by the cellat .${\mathbf{25}}^{\mathbf{°}}\mathbf{\text{C}}$

Short answer

The initial voltage generated by the cell at${25}^{°}\text{C}$is$\text{E}=0.381\text{\hspace{0.17em}V}$.

Step-by-step solution

The cell potential in volts  

The cell potential is given by

$\mathbf{\text{E}}\mathbf{=}{\mathbf{\text{E}}}^{\mathbf{°}}\mathbf{-}\frac{\mathbf{0}\mathbf{.059}\mathbf{\text{V}}}{\mathbf{\text{n}}}\mathbf{\times }\mathbf{log}\mathbf{\text{Q}}$

where $\text{E}$is the cell potential in volts ${\mathbf{\text{E}}}^{\mathbf{°}}$ is the standard cell potential in volts nis the number of moles of electrons transferred in the redox reaction Q is the reaction quotient.

 Step 2: The cell's initial voltage

The cell's initial voltage is measured in volts .${25}^{°}\text{C}$

$\begin{array}{l}{{\text{Pb}}^{2+}}_{\left(aq\right)}+2{\text{Cr}}^{2+}\left(aq\right)\to {\text{Pb}}_{\left(s\right)}+2{{\text{Cr}}^{3+}}_{\left(aq\right)}\\ \left[{{\text{Pb}}^{2+}}_{\left(aq\right)}\right]=0.15\text{M}{[{\text{Cr}}^2+}_{\left(aq\right)}]\\ =0.20\text{M}\left[{{\text{Cr}}^{3+}}_{\left(aq\right)}\right]\\ =0.0030\text{M}\end{array}$

Calculate the cell potential in volts using the formula 

The cell potential in volts is given by

$\text{E}={\text{E}}^{°}-\frac{0.059\text{V}}{\text{n}}\times \log \text{Q}$

where $\text{E}$ is the cell potential in volts ${\text{E}}^{°}$ is the standard cell potential in volts is the

number of moles of electrons transferred in the redox reaction $Q$ is the reaction quotient.

The anode undergoes oxidation, while the cathode undergoes reduction

Calculate standard cell potential for each component using standard reduction potentials.

Calculate the standard cell potential for each component using standard reduction potentials, as follows:

$\begin{array}{l}{\text{Pb}}^{2+}+2{\text{e}}^{-}\to \text{Pb}\\ {\text{E}}^{°}-1=-0.1263\text{V}\\ {\text{Cr}}^{2+}\to {\text{Cr}}^{3+}+{\text{e}}^{-}\\ {\text{E}}^{°}2=+0.424\text{V}\end{array}$

The cell's standard potential will be:

$\begin{array}{l}{\text{E}}_{{}^1}^{\circ }+{\text{E}}_2^{\circ }=-0.1263\text{V}+0.424\text{V}\\ {\text{E}}^{°}=0.298\text{V}\end{array}$

 Step 5: The final concentration

Multiply the starting concentration by 2 to get the final concentration.

$\text{E}={\text{E}}^{°}-\frac{0.059\text{V}}{\text{n}}\times \log Q$

Plug in the values for the starting concentration, and multiply the initial concentration by 2 if there are 2 moles of an ion.

$\begin{array}{l}\text{E}=0.298\text{V}-\frac{0.059\text{V}}{2}\times \log \frac{{\left(0.0030\text{M}\right)}^2}{\left(0.15\text{M}\right){\left(0.20\text{M}\right)}^2}\\ \text{E}=0.381\text{\hspace{0.17em}V}\end{array}$